zoukankan      html  css  js  c++  java
  • 数论基础算法总结(python版)

    /*
        Author: wsnpyo
        Update Date: 2014-11-16
    Algorithm: 快速幂/Fermat, Solovay_Stassen, Miller-Rabin素性检验/Exgcd非递归版/中国剩余定理 */ import random def QuickPower(a, n, p): # 快速幂算法 tmp = a ret = 1 while(n > 0): if (n&1): ret = (ret * tmp) % p tmp = (tmp * tmp) % p n>>=1 return ret def Jacobi(n, m): # calc Jacobi(n/m) n = n%m if n == 0: return 0 Jacobi2 = 1 if not (n&1): # 若有n为偶数, 计算Jacobi2 = Jacobi(2/m)^(s) 其中n = 2^s*t t为奇数 k = (-1)**(((m**2-1)//8)&1) while not (n&1): Jacobi2 *= k n >>= 1 if n == 1: return Jacobi2 return Jacobi2 * (-1)**(((m-1)//2*(n-1)//2)&1) * Jacobi(m%n, n) def Exgcd(r0, r1): # calc ax+by = gcd(a, b) return x x0, y0 = 1, 0 x1, y1 = 0, 1 x, y = r0, r1 r = r0 % r1 q = r0 // r1 while r: x, y = x0 - q * x1, y0 - q * y1 x0, y0 = x1, y1 x1, y1 = x, y r0 = r1 r1 = r r = r0 % r1 q = r0 // r1 return x def Fermat(x, T): # Fermat素性判定 if x < 2: return False if x <= 3: return True if x%2 == 0 or x%3 == 0: return False for i in range(T): ran = random.randint(2, x-2) # 随机取[2, x-2]的一个整数 if QuickPower(ran, x-1, x) != 1: return False return True def Solovay_Stassen(x, T): # Solovay_Stassen素性判定 if x < 2: return False if x <= 3: return True if x%2 == 0 or x%3 == 0: return False for i in range(T): # 随机选择T个整数 ran = random.randint(2, x-2) r = QuickPower(ran, (x-1)//2, x) if r != 1 and r != x-1: return False if r == x-1: r = -1 if r != Jacobi(ran, x): return False return True def MillerRabin(x, ran): # x-1 = 2^s*t tx = x-1 s2 = tx&(~tx+1) # 取出最后一位以1开头的二进制 即2^s r = QuickPower(ran, tx//s2, x) if r == 1 or r == tx: return True while s2>1: # 从2^s -> 2^1 循环s次 r = (r*r)%x if r == 1: return False if r == tx: return True s2 >>= 1 return False def MillerRabin_init(x, T): #Miller-Rabin素性判定 if x < 2: return False if x <= 3: return True if x%2 == 0 or x%3 == 0: return False for i in range(T): # 随机选择T个整数 ran = random.randint(2, x-2) if not MillerRabin(x, ran): return False return True def CRT(b, m, n): # calc x = b[] % m[] M = 1 for i in range(n): M *= m[i] ans = 0 for i in range(n): ans += b[i] * M // m[i] * Exgcd(M//m[i], m[i]) return ans%M

    以上作为半个学期来数论学习的一个小结,也许以后难以再系统的学习数论了。略伤感咿

      —— 多谢信息安全数学基础的老师

  • 相关阅读:
    bash实现多进程运行
    Erlang实现进程池
    Apache 和nginx支持跨域访问
    thinkphp 发送邮件
    判断PC和移动端 判断移动端系统
    check 选择框checked属性读取不到
    php使用PDO,并连接SQL
    SQL2005:由于目标机器积极拒绝,无法连接
    frozenui 移动端ui
    纯jquery 滚动评论
  • 原文地址:https://www.cnblogs.com/Mathics/p/4101467.html
Copyright © 2011-2022 走看看